First, let us consider a complex differential equation: \dot{z}=f(z,t) \tag{1} If we let z=x+yi and f(z,t)=f(x+yi,t)=g(x,y,t)+i \cdot h(x,y,t), then the equation corresponds to: \begin{aligned} \dot{x}&=g(x,y,t) \\ \dot{y}&=h(x,y,t) \end{aligned} This shows that, to a certain extent, a system of two-variable differential equations is equivalent to a single complex differential equation.
We can look at an example. Solve the system of equations: \dot{x}=x^2-y^2, \dot{y}=2xy, with initial values at t=0, x=1, y=1.
If we let z=x+yi, then the equation is actually equivalent to \dot{z}=z^2, with the initial condition t=0, z=1+i. Thus, using the method for solving a single-variable differential equation, we obtain: z=-\frac{1}{t+C}, and according to the initial condition, we find C=1/2 (i-1), which gives: x+yi=z=-\frac{1}{t+1/2 (i-1)}=\frac{1/2-t}{(1/2-t)^2+(1/2)^2}+\frac{1/2 i}{(1/2-t)^2+(1/2)^2}
According to the principle of equality of complex numbers, we obtain: \begin{aligned} x&=\frac{1/2-t}{(1/2-t)^2+(1/2)^2} \\ y&=\frac{1/2}{(1/2-t)^2+(1/2)^2} \end{aligned}
As can be seen, the method is very efficient; this is an important way to transform the unfamiliar into the familiar. An obvious advantage of converting a system of two-variable real differential equations into a single-variable complex differential equation is the reduction of variables. Rather than calling it a conversion between different forms, it is better described as a "method of elimination." Furthermore, we have previously discussed the equivalence between complex numbers and two-dimensional vectors. Therefore, differential equations involving two-dimensional vectors can be solved by treating the vectors as "numbers" (complex numbers), provided that the vector equation does not involve the "magnitude" (norm). Solving complex differential equations is, to a large extent, the same as solving real differential equations. Many equations proposed in practical problems (especially in physics and mechanics) are ideal two-dimensional vector equations. Since vectors are equivalent to complex numbers, why not use the form of complex differential equations to solve them quickly and conveniently?
Below is a practical application in kinematics:
A projectile is fired from the ground with an initial velocity v_0 at an angle \theta to the ground. The air resistance is q times the velocity (q is a constant). Find the trajectory of the projectile.
Let the mass of the projectile be m, the initial velocity be \vec{v}_0, the gravitational acceleration be \vec{g}, and the trajectory be \vec{r}. We can list the equation: \ddot{\vec{r}}=-\frac{q}{m}\dot{\vec{r}}+\vec{g}. Let -\frac{q}{m}=k, then: \ddot{\vec{r}}=k\dot{\vec{r}}+\vec{g} Integrating once and substituting the initial conditions, we get: \dot{\vec{r}}=k\vec{r}+\vec{g}t+\vec{v}_0 Since vectors are equivalent to complex numbers, if we treat the vector as a complex function, the above expression is a first-order differential equation. Using the knowledge of solving first-order differential equations, we obtain: \vec{r}=\left(\frac{\vec{g}}{k^2}+\frac{\vec{v}_0}{k}\right)(e^{kt}-1)-\frac{\vec{g}t}{k}
Let \vec{r}=x+yi, \vec{g}=-gi, \vec{v}_0=v_0 \cos\theta+i \cdot v_0 \sin\theta (which is equivalent to the orthogonal decomposition of the vectors). Substituting these into the above equation and applying the condition for equality of complex numbers, we obtain: \begin{aligned} x&=\left(\frac{v_0 \cos\theta}{k}\right)(e^{kt}-1) \\ y&=\left(\frac{v_0 \sin\theta}{k}-\frac{g}{k^2}\right)(e^{kt}-1)+\frac{g t}{k} \\ k&=q/m \end{aligned}
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